How to prove this?

Question

Let G be a finite group and H a proper subgroup such that |G| doesn't divide |G/H|!. Prove that H contains a nontrivial normal subgroup of G.

Answer 1

Let f be the mapping from G to the group of permutations of the left cosets of G in H defined by f(x) = psubx, where psubx is the permutation of left cosets defined by

psubx(yH)=xyH.

Then obviously (after you think about it for an hour), f is a homomorphism. (psubx * psuby = psubxy).

Therefore the kernel of f is a normal subgroup of G. But if x is in the kernel of f then psubx is the identity permutation, so psubx(H)=H. So xH=H. So x lies in H.

Now suppose that kernel of f is trivial. Then G has order dividing [G:H]!, contradiction. Therefore kefnel of f is not trivial.

Note: I learned this decades ago in a class. The reference is lost in the mists of time.